let f be continuous
for any x>1 and [] denoting the greatest integer function
prove that
\int_{1}^{x}(+1)f(u)du =2\sum_{i=1}^{[x]}{i\int_{i}^{x}{f(u)du}}
9its a matter of splitting the integrations
in RHS for each integration btw to integers for a specific ni its of form 2 Σ i = n(n+1)
which is wat in LHS
11wat is trivial bhool gaya[2]
ur answer follows
\int_{1}^{2}{2f(u)du}+\int_{2}^{3}{6(f(u)du}+........................\int_{x-1}^{x}{x(x-1)f(u)du}
2(\int_{1}^{2}{f(u)du}+\int_{2}^{3}{3f(u)du}................\int_{x-1}^{x}{\frac{x(x-1)}{2}f(u)du}
2\sum_{i=1}^{[x]}i{\int_{i}^{x}{f(u)du}}=2(\int_{1}^{x}{f(u)du}+2\int_{2}^{x}{f(u)du}..........[x]\int_{[x]}^{x}{f(u)du})
which is equal to LHS
as[x]=x-1 if x is not integer
1hey manipal
check your limits in LHS and RHS ...
and celestine where is your doubt in LHS ..?
11rohan bhai maine open kar diya
i m,eant
x ko break kar liya tha in the LHS
1but how can u write [x]=x-1 when x is not an integer ?
and , in the LHS we have terms having upper limit verying but in the RHS every term has an upper limit x..
9rohan this is very trivial due u need step by step method ??
i saw that geometry q posted as a starter ( looks like a toughie to me )
i think ill better spend time on that